Thinking a bit more about it, this problem already pops up if the bottom face has either:
- three edge pieces that share a color, with that color showing on the side.
- two pairs of edge pieces that share a color, with the twice shared colors showing on the side.
The first isn’t uncommon; if you pick three edges at random, the probability that they share a color is 1 * 6/11 * 2/10 = 6/55, and there are four ways to pick three edges from the bottom face, for (I think) a total probability of 24/55 - 4 * 6/55 * 1/9 (the probability that all four share the same color), or a probability of over 1/3 (if that seems high, consider the following: because of the pigeonhole principle, between the edges of any face of the cube there always is at least a pair sharing a color). That means that, even ignoring the second possibility, there’s at least a 1:24 probability of seeing this problem on a random cube.
- three edge pieces that share a color, with that color showing on the side.
- two pairs of edge pieces that share a color, with the twice shared colors showing on the side.
The first isn’t uncommon; if you pick three edges at random, the probability that they share a color is 1 * 6/11 * 2/10 = 6/55, and there are four ways to pick three edges from the bottom face, for (I think) a total probability of 24/55 - 4 * 6/55 * 1/9 (the probability that all four share the same color), or a probability of over 1/3 (if that seems high, consider the following: because of the pigeonhole principle, between the edges of any face of the cube there always is at least a pair sharing a color). That means that, even ignoring the second possibility, there’s at least a 1:24 probability of seeing this problem on a random cube.